Koch Proof
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Proof that the Koch Snow Flake constitutes of an infinite line surrounding a finite area.
| Generation n | Surface of single triangle | Amount of triangles generation n | Total surface added generation n |
| 0 | A | 1 | A |
| 1 | 1/9 A | 3*40 | 3*40 * 1/9 A |
| 2 | 1/92 A | 3*41 | 3*41 * 1/92 A |
| 3 | 1/93 A | 3*42 | 3*42 * 1/93 A |
| 4 | 1/94 A | 3*43 | 3*43 * 1/94 A |
| n | 1/9n A | 3*4n-1 | 3*4n-1 * 1/9n A |
| Total surface up to generation n |
| S=A + 3*40 * 1/91 A + 3*41 * 1/92 A + 3*42 * 1/93 A + ...+ 3*4n-1 * 1/9n A |
| Both sides times 4 * 1/9 |
| S * 4/9 = (4/9) * A + 3*41 * 1/92 A + 3*42 * 1/93 A +...+ 3*4n-1 * 1/9n A + 3*4n * 1/9n+1 A |
| Subtracting S- 4/9 S to lose the infinite tail |
| S- 4/9 S = A + 3*40/91 A - 4/9 A - 3*4n/9n+1 A 5/9 S = 5/9 A + 3/9 A - 3*4n/9n+1 A = 8/9 A - 3*4n/9n+1 A S = 8/5 A - 9/5 * 3 * 4n/9n+1 = 8/5 A - 3/5 * (4/9)n |
Conclusion: The final surface of the Koch Snow Flake will be 1 3/5 times the surface of the parent triangle after an infinite number of generations.
1.2 The outer line L
| Every generation will cause an increase in length of the curve of 4/3 times a segment of the previous generation. Let the length of the side of the parent triangle be R. |
| L = 3R + 3* 40/31 R + 3*41/32 R+...+ 3*4n-1/3n R L =3R + (4/3)0 R + (4/3)1 R + (4/3)2 R +...+ (4/3)n-1 R |
| Both sides times 4/3 |
| 4/3 L = 4R + (4/3)1 R + (4/3)2 R +...+ (4/3)n-1 R + (4/3)n R |
| Subtracting 4/3 L - L to lose the infinite tail |
| 4/3 L - L = 1/3 L = 4R + (4/3)n R - 3R - (4/3)0 R = (4/3)nR L = 4 * (4/3)n-1 |
Conclusion: The final length of the Koch Snow Flake will be infinite after an infinite number of generations.
